When teaching functions in polar coordinates, the nearly universal practice is to consider functions of the form $r = f(\theta)$. I think I have never seen any examples in which $\theta$ is expressed as a function of $r$. My intuition tells me that the usual setup is more ``natural'' in some sense than the latter, but I am not sure whether that is just because it's what I'm used to, or if I am responding to some intrinsic difference between the two cases. Is there any good reason why, for instance, it seems more natural to write $r = \sqrt{\theta}$ than to write $\theta = r^2$, despite the fact that they are (obviously) equivalent?
Edited to add: Here is an example of what I mean. Consider the function $\theta = \cos (r)$. It has a very cool graph:
Intuitively, if we think of $r$ as the independent variable, and ask how $\theta$ depends on $r$, the description is quite simple: as we move outwards from the origin, the angle oscillates back and forth between a minimum angle of $-1$ radian and a maximum angle of $1$ radian. This is a perfectly reasonable function to think about, graph, and study. And yet functions like this are (as far as I can tell) completely absent both from textbooks and from usual instruction. I find this strange, and am wondering if anybody has any thoughts on why.
Addendum-2 added to respond to the OP's editing of their original posting, re the function $\theta = \cos(r)$.
Addendum added to respond to a comment question.
Because a function must map each element in its domain to only a single element in its range.
For example, consider the function $r = \cos(\theta)$.
Then, for each angle $\theta$, there is a specific value $r$. However, the function does not have an inverse. For example if you try to express $\theta = g(r)$, then what is $g(1/2)$? Even if you restrict the range to $0 \leq \theta < 2\pi$, for example, would $g(1/2)$ equal $\pi/3$ or $5\pi/3$?
One approach, is to force the inverse function $g(r)$ to exist by requiring that the range of $\theta$ be limited to $0 \leq \theta \leq \pi.$ In fact, this is exactly what is done with the Arccos function, and for this very reason.
Normally, when you have a function like $r = f(\theta)$, you want $\theta$ to at least take on the values in some half-open interval of width $(2\pi)$, and perhaps allow $\theta$ to take on the value of any real number.
By specifying $r = f(\theta)$, instead of attempting to specify $\theta = g(r)$, you don't have to worry whether a specific value of $r$ is associated with more than one value of $\theta$.
Edit
Certainly, you can contrive functions, like $r = 2\theta + 4$, where the considerations of the previous section are a non-issue. However, normally, a function like $r = f(\theta)$ will be used when the function $f$ is performing some trigonometric action, such as some interaction between the sine and cosine functions.
Addendum
Responding to the comment question of Akiva Weinberger.
First of all, $~\displaystyle r = \sqrt{\theta} ~$ is only defined on $~\theta \geq 0~$, while $\displaystyle ~ r = \sqrt{\theta + 2\pi} ~$ is only defined on $~\theta \geq -2\pi.~$
Then, $\displaystyle ~\sqrt{\theta} ~$ is not equal to $\displaystyle ~\sqrt{\theta + 2\pi}. ~$
Second of all, the $~r = f(\theta)~$ approach is normally only used when $~f~$ relates to trigonometry functions.
So, consider the alternate of function of $~r = f(\theta) = \sqrt{|\cos \theta|}~$.
This function does not present any difficulty. It is merely a function that is not injective. The situation analogizes to setting $y = f(x) = |x| ~: x \in \Bbb{R}.$ You wouldn't be able to construct the inverse function $x = g(y)$, because (for example) $g(5)$ would ambiguously refer to $+5,$ or $-5$.
In the same way, when $~r = f(\theta) = \sqrt{|\cos \theta|}~$, $f(\theta) = f(\theta + 2\pi)$.
Notice what happens if you try to construct the alternative corresponding mapping of $\theta = g(r).$ What would $g(1)$ map to, $\theta = 0, \theta = \pi, \theta = 2\pi,$ or what?
You could not use the corresponding mapping to construct an inverse function because the original mapping was not injective. This is typical for $r = f(\theta)$ functions that involve trigonometric operations.
In a similar vein, consider the polar function $r = \cos^2(\theta) + \sin^2(\theta).$
Addendum-2
Reacting to the OP's editing of their original posting, re $\theta = \cos(r)$.
This is a function that I never thought of, because of my blind spot that the magnitude should be a function of the angle.